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Algebra Difficulty 6.1 National olympiad Prove it

Example 49 Non-negative real numbers a,b,ca, b, c satisfy (a+b)(b+c)(c+a)=2(a+b)(b+c)(c+a)=2. Prove:
(a2+bc)(b2+ca)(c2+ab)⩽1\left(a^{2}+b c\right)\left(b^{2}+c a\right)\left(c^{2}+a b\right) \leqslant 1

Solution

Prove (a2+bc)(b2+ca)(c2+ab)⩽1\left(a^{2}+b c\right)\left(b^{2}+c a\right)\left(c^{2}+a b\right) \leqslant 1
⇔(a+b)2(a+c)2(b+c)2⩾4(a2+bc)(b2+ca)(c2+ab)⇔(a−b)2(a−c)2(b−c)2+∑sym (4a3b2c+43a2b2c2)⩾0.\begin{array}{l} \Leftrightarrow(a+b)^{2}(a+c)^{2}(b+c)^{2} \geqslant 4\left(a^{2}+b c\right)\left(b^{2}+c a\right)\left(c^{2}+a b\right) \\ \Leftrightarrow(a-b)^{2}(a-c)^{2}(b-c)^{2}+\sum_{\text {sym }}\left(4 a^{3} b^{2} c+\frac{4}{3} a^{2} b^{2} c^{2}\right) \geqslant 0 . \end{array}

The above expression is obviously true, hence the original inequality holds.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.